Relationship between Connection and Material Derivative Suppose $D\subset \Bbb R^3$ contains a fluid and that $f : D\times \mathbb{R}\to \mathbb{R}$  is a time dependent function defined on the fluid region. In that case, the material derivative is defined by
$$\dfrac{D}{Dt}f(a, t) = \dfrac{\partial f}{\partial t}(a, t) + (\mathbf{u}\cdot \nabla)f(a, t)$$
Where $\mathbf{u}\cdot \nabla$ is the operator defined on scalar function $g$ by
$$(\mathbf{u}\cdot \nabla )g = \mathbf{u}\cdot (\nabla g) = D_{\mathbf{u}} g.$$
That is the directional derivative along $\mathbf{u}$ of the function $g$. On vector fields it is defined componentwise, that is, if $\mathbf{v} = (v_1,v_2,v_3)$ then
$$(\mathbf{u}\cdot \nabla)\mathbf{v} = ((\mathbf{u}\cdot \nabla)v_1, (\mathbf{u}\cdot \nabla)v_2, (\mathbf{u}\cdot \nabla)v_3) = (D_{\mathbf{u}}v_1, D_{\mathbf{u}}v_2, D_{\mathbf{u}}v_3).$$
But that latter thing is clearly the Covariant Derivative of $\mathbf{v}$ along $\mathbf{u}$ when we consider the Levi-Civita Connection on $\mathbb{R}^3$ with the usual flat metric tensor, that is
$$(\mathbf{u}\cdot \nabla)\mathbf{v} = \nabla_{\mathbf{u}}\mathbf{v}.$$
Now, is this conclusion right? Can we really write the material derivative as
$$\dfrac{D}{Dt}\mathbf{v}(a, t) = \dfrac{\partial \mathbf{v}}{\partial t}(a, t) + \nabla_{\mathbf{u}}\mathbf{v}(a, t),$$
and if it's right, is there some usefulness in this relationship? I mean, I don't know that much of connections and how they can be used on Physics, but I know they are usefull. In that case, writing the material derivative in terms of a connection gives some advantage? Would it make sense if $\nabla$ were another connection other than the Levi-Civita connection?
 A: Let there be given an $n$-dimensional manifold $(M,\nabla)$ endowed with a connection $\nabla$. [In particular, we do not assume that the manifold $M$ is equipped with a metric tensor.] Let there be given a curve $\gamma:\mathbb{R}\to M$. Here the reader should think of $\mathbb{R}$ and $M$ as time and space, respectively. 


*

*If $f: M\times \mathbb{R}\to \mathbb{R}$ is a time-dependent scalar on $M$, then the material/total derivative is
$$\tag{1} d_t f = \partial_t f + \dot{\gamma}^i \partial_i f.$$
In particular, the material/total derivative $d_t f$ of a scalar $f$ is independent of the connection $\nabla$.

*If $V$ is a time-dependent vector field on $M$, then the material/total derivative is
$$\tag{2} d_t V = \partial_t V + \dot{\gamma}^i \nabla_i V.$$
A: NOTE ADDED. When I answered the question, I completely misinterpreted it. I thought it was related to the connection and the covariant derivative used in General Relativity. It's not the case at all. I decided to keep the answer anyway,   because perhaps someone with the same interpretation as I had will be directed here  by search engines, as I was.  What follows is my original answer, which starts by providing my totally incorrect, but interesting interpretation of the question. 
The OP asked three questions. First, he asked if it makes sense to say that the covariant derivative along the four-velocity $\mathbf{u}$ is the material derivative. Second, he asked if such a relationship is useful. Third, given that the covariant derivative is a special kind of connection, he asked if this relationship still hold and is useful for other form of connections.  
The answer to the first question is that the covariant derivative in  the direction of the four-velocity (i.e., $\mathbf{u}^{\alpha}\nabla_{\alpha}$, where $\mathbf{u}$ is the four-velocity of the particle or fluid element and $\nabla$ is the covariant derivative)  is the generalization in curved space-time of the material derivative. 
We cannot say that this covariant derivative (in the direction of velocity) is the material derivative.  We have an analogous situation with the relativistic four-velocity. We cannot say that the ordinary velocity is the same thing as the relativistic four-velocity, but it is correct to say that the relativistic four-velocity is the generalization of the ordinary velocity. 
Let us see how the covariant derivative in the direction of the velocity reduces to the material derivative when we are in an inertial frame of reference and in the low velocity limit. First, in an inertial frame of reference, we have that the covariant derivative reduces to the ordinary partial derivative.  So, in an inertial frame, we have $$\mathbf{u}^{\alpha}\nabla_{\alpha} = \mathbf{u}^{\alpha}\partial_{\alpha}.$$ 
Note that $\nabla_i$, where $i$ is one of the components $x$, $y$ or $z$ is often used to denote the usual partial derivative, but we prefer to use $\partial_i$ to denote the partial derivative to avoid a confusion with the covariant derivative. They are only identical in an inertial coordinate system. 
Second, the four-velocity (which, by definition, as lenght $-1$ in the Minskowki metric) is $$\mathbf{u} = \frac{(1, v^x, v^y, v^z)}{(1 - v^2)^{1/2}},$$ where we used units such that the speed of light is $c = 1$. As a parenthesis, we see that, in the low velocity limit $v \approx 0$, we have $\gamma =  (1 - v^2)^{-1/2} \approx 1$ and the spatial component $\gamma (v^x, v^y, v^z)$ of the four-velocity is close to the ordinary velocity. We now show that, in a similar way, $\mathbf{u}^{\alpha}\partial_{\alpha}$ reduces to the material derivative in the low velocity limit. We use that $$v^x = \frac{dx}{dt}, \ \ v^y = \frac{dy}{dt}, \ \ v^z = \frac{dz}{dt}.$$ In the low velocity limit $v \approx 0$, we accept $(1 - v^2)^{1/2} \approx 1$ and, ignoring the small contribution of higher terms in $v$, we obtain $$\mathbf{u}^{\alpha}\partial_{\alpha} = \partial_t + \frac{dx}{dt} \partial_x + \frac{dy}{dt} \partial_y + \frac{dz}{dt} \partial_z,$$ which is the definition of the material derivative. 
Regarding the second question, this relationship between covariant derivative and material derivative is, among other things, useful to see  how the conservation equation $$\nabla_\mu T^{\mu\nu} = 0$$ of General Relativity, where $T$ is the stress-energy tensor, relates to the standard equations of Newtonian fluid dynamics. We use the principle that the laws must remain the same under a change of coordinate system and that we can always find a coordinate systems that is locally inertial.  For example, this is useful to obtain the typical Euler equations under adequate assumptions. For the conservation of energy component, in an inertial frame of reference, we take the projection on the four-velocity $\mathbf{u}$ to get the timelike component of the GR equation and then take the low velocity limit. See the details in "Relativistic Fluid Dynamics" http://mathreview.uwaterloo.ca/archive/voli/2/, sections 2.1 and 3.3. 
Regarding the third question, I am not sure, but it seems that the same kind of strategy that we used to relate the GR equations to the Newtonian equations might apply in the general case. In the case of the Levi-Civita connection, the strategy is based on the postulated principle that the laws are the same in any system of coordinates. We would most likely need an extension of this principle. Some thing to think about. I am not sure. It is also not clear to me that such a level of generality is physically meaningful. 
